Question Number 174336 by cortano1 last updated on 30/Jul/22
$$\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\:\frac{\mid{x}−\mathrm{2}\mid}{{x}^{\mathrm{2}} −\mathrm{4}{x}}\:{dx}\:=? \\ $$
Answered by Mathspace last updated on 30/Jul/22
![I=−∫_1 ^2 ((x−2)/(x^2 −4x))dx+∫_2 ^3 ((x−2)/(x^2 −4x))dx f(x)=((x−2)/(x(x−4)))=(a/x)+(b/(x−4)) a=((−2)/(−4))=(1/2) and b=((4−2)/4)=(1/2) ⇒∫_1 ^2 f(x)dx=(1/2)∫_1 ^2 ((1/x)+(1/(x−4)))dx =(1/2)[ln∣x∣+ln∣x−4∣]_1 ^2 =(1/2){2ln2−ln3} ∫_2 ^3 ((x−2)/(x^2 −4x))dx=(1/2)∫_2 ^3 ((1/x)+(1/(x−4)))dx =(1/2)[ln∣x∣+ln∣x−4∣]_2 ^3 =(1/2){ln3−2ln2} ⇒ ∫_1 ^3 ((∣x−2∣)/(x^2 −4x))dx=−(1/2){2ln2−ln3} +(1/2){ln3−2ln2} =−ln2+(1/2)ln3+(1/2)ln3−ln2 =ln3−2ln2 .](https://www.tinkutara.com/question/Q174342.png)
$${I}=−\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx}+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx} \\ $$$${f}\left({x}\right)=\frac{{x}−\mathrm{2}}{{x}\left({x}−\mathrm{4}\right)}=\frac{{a}}{{x}}+\frac{{b}}{{x}−\mathrm{4}} \\ $$$${a}=\frac{−\mathrm{2}}{−\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{b}=\frac{\mathrm{4}−\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\mathrm{2}} {f}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{4}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid{x}\mid+{ln}\mid{x}−\mathrm{4}\mid\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}{ln}\mathrm{2}−{ln}\mathrm{3}\right\} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{2}} ^{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{4}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid{x}\mid+{ln}\mid{x}−\mathrm{4}\mid\right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\mathrm{3}−\mathrm{2}{ln}\mathrm{2}\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \frac{\mid{x}−\mathrm{2}\mid}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}{ln}\mathrm{2}−{ln}\mathrm{3}\right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\mathrm{3}−\mathrm{2}{ln}\mathrm{2}\right\} \\ $$$$=−{ln}\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}−{ln}\mathrm{2} \\ $$$$={ln}\mathrm{3}−\mathrm{2}{ln}\mathrm{2}\:. \\ $$